Yes, this is true. It is equivalent to the fundamental theorem of algebra which says that $\mathbb{C}=\mathbb{R}(i)$ is algebraically closed.

If $p \in \mathbb{R}[x]$ is a monic polynomial, then it is a product of linear factors $x-a$ with $a \in \mathbb{C}$. The complex conjugation leaves $p$ invariant, hence acts on these roots $a$. If it leaves $a$ invariant, this means $a \in \mathbb{R}$ and we keep $x-a$. If not, we expand $(x-a)(x-\overline{a})=x^2-(a + \overline{a})x + a\overline{a} \in \mathbb{R}[x]$.

For example, $p=x^3+3x^2+x+3$ has the roots $i,-i,-3$, so that it factors as $(x^2+1)(x+3)$.

So, if polynomial with real coefficients has any complex root, that it has quadratic "polynomial factor". If it doesn't have any complex root, then it also has such "factor" (or it's polynomial of first degree). Factor it out and repeat.